# A container with a volume of 14 L contains a gas with a temperature of 160^o C. If the temperature of the gas changes to 320 ^o K without any change in pressure, what must the container's new volume be?

Mar 23, 2016

${V}_{2} = 28 L$ the volume will double
${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$ Note this a special case of the combined Charles and Boyle Law:
$\frac{{V}_{1} {P}_{1}}{T} _ 1 = \frac{{V}_{2} {P}_{2}}{T} _ 2$, with ${P}_{1} = {P}_{2}$
$\frac{14 L}{{160}^{o} C} = \frac{{V}_{2}}{{320}^{o} C}$
${V}_{2} = 28 L$ the volume will double